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I've recently finished one semester in differential topology (with Milnor's Topology from the Differentiable Viewpoint) and my first semester of algebraic topology. I believe I understand Milnor's definition of what it means for two manifolds to be cobordant: roughly speaking, two n-manifolds are cobordant if there exists an (n+1)-manifold whose boundary is the disjoint union of the two original n-manifolds.

My question is how does one go from this definition to a cohomology theory? I'm not exactly sure this is the right question to ask, so please provide any insight you can. I naively understand spectra and Brown representability, so you can answer in those terms if you'd like (I'm sure these concepts have something to do with the answer, but I can't quite piece it together). The wiki page for complex cobordism may have the answer, but again, I can't decipher it.

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    $\begingroup$ en.wikipedia.org/wiki/… has a rough description of how to construct the cohomology theory associated to cobordism. It describes what elements of the groups are, and when they are equivalent, in a very concrete way. What can't you decipher? (Usually, «explain X to me» is not a good way to use this site: it is immensely better if you ask something specific) $\endgroup$ Commented Jan 5, 2012 at 17:21

4 Answers 4

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It is a bit easier to describe the homology theory associated with cobordism; let's say unoriented cobordism here. Then one gets a homology theory $MO_*$ such that elements of $MO_*(X)$ can be described as an $n$-manifold $M$ together with a map $f: M \to X$ (this would live in degree $n$). Two maps $f: M \to X, g: N \to X$ have the same class in $MO_*(X)$ if there exists a manifold-with-boundary $P$ of dimension $n+1$ together with a map $H: P \to X$ such that $\partial P = M \sqcup N$ and the restrictions of $H$ to $M, N$ are just $f, g$ respectively. In this way, you can build a homology theory out of the ways in which manifolds map to $X$.

The associated spectrum, by the Thom-Pontryagin construction, is the Thom spectrum $MO$ obtained as follows. Take the classifying space $BO_n$ for the orthogonal group $O(n)$; on it is a universal $n$-dimensional vector bundle $\zeta_n$. The Thom space $MO(n)$ of $BO(n)$ can now be defined. Because of the natural maps $$BO(n) \to BO(n+1)$$ pulling back $\zeta_{n+1}$ to $\zeta_n \oplus 1$, one gets maps $$MO(n) \wedge S^1 \to MO(n+1)$$ (because $\zeta_n\oplus1$ has Thom space $MO(n) \wedge S^1$). In this way, one gets the spectrum (which, as I've defined, is not an $\Omega$-spectrum, so some would call it a prespectrum) $MO$, which represents unoriented bordism.

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  • $\begingroup$ Thanks, this explanation is the sort of thing I was looking for. $\endgroup$
    – mcat
    Commented Jan 6, 2012 at 17:37
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There is a fantastic set of notes by Haynes Miller

Haynes Miller - Notes on Cobordism

from a 1994 class of his, that details this and probably much more than you would ever want to know! In particular the first 15 pages or so show the relation between bordism and the spectrum $MO$.

There is also the classic - 'Notes on Cobordism Theory' by Stong, but I would definitely start with Haynes Miller's notes.

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  • $\begingroup$ Heh. I just found Miller's notes independently and, after being impressed by how nice they were, was about to add a link to them... $\endgroup$ Commented Jan 6, 2012 at 3:12
  • $\begingroup$ Someone really needs to make a listing of all the nice algebraic topology notes on the Internet (COCTALOS, Miller's notes on cobordism and Rezk's notes on the Hopkins-Miller theorem for a start)! $\endgroup$
    – Juan S
    Commented Jan 6, 2012 at 4:56
  • $\begingroup$ These look great. Thanks! $\endgroup$
    – mcat
    Commented Jan 6, 2012 at 17:36
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The relationship between cobordism and generalized cohomology was first revealed by Atiyah, in his paper Bordism and cobordism (Proc. Camb. Phil. Soc. 57, 200-208 (1961))

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The OP asked about a description of the cohomology theory of Bordism, which no one above did in an explicit way.

Given a manifold $X$, an element of $\Omega^*(X)$ is represented by a smooth proper map $M \xrightarrow{i} X$ together with the data of the normal bundle $\nu(i)$ of $i$. $(M_1,i_1)$ and $(M_2,i_2)$ are equivalent if there is a (smooth proper) map $M \xrightarrow{i} X \times I$ such that $\nu(i)$ restricts to $\nu(i_1)$ and $\nu(i_2)$ at the $\partial M =M_1 \sqcup M_2$.

This cohomology class will be in degree $deg(X)-deg(M_1)$. For example when $X$ is a point, one gets cohomology in the negative degrees. This corresponds to the fact that for any spectrum $E$-homology of a point will be $\pi_*(E)$ and the $E$-cohomology of a point will be $\pi_{-*}(E)$. Its easy to see from this explicit construction that $MO^*(pt)$ is exactly $MO_{-*}(pt)$ as it should be. Checking that this is the cohomology theory dual to bordism homology in the sense of Akhil Matthew's answer can by done by looking at the details of the Pontryagin Thom construction.

For more information see the paper by Quillen: Elementary Proofs Of Some Results Of Cobordism Theory Using Steenrod Operations where this was introduced.

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  • $\begingroup$ By smooth do you mean a submersion? $\endgroup$ Commented Oct 27, 2018 at 9:22

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